91Ó°ÊÓ

Here $\frac{\mathrm{dx}}{\mathrm{dt}}=1+\mathrm{t}\sin \left(\mathrm{tx}\right),\mathrm{x}\left(0\right)=0$,

For $\mathrm{ξ}=0.01$, x = 0, , c = 0, M = 10, h = 1

$$\begin{gathered} \mathrm{F}=\mathrm{f}(\mathrm{t},\mathrm{x}) \\ =1+\mathrm{tsin}\left(\mathrm{tx}\right) \\ \mathrm{G}=\mathrm{f}(\mathrm{t}+\mathrm{h},\mathrm{x}+\mathrm{hF}) \\ =1+(\mathrm{t}+\mathrm{h})\sin \left(\right(\mathrm{t}+\mathrm{h}\left)\right(\mathrm{x}+\mathrm{hF}\left)\right) \end{gathered}$$

Apply initial points $\mathrm{t}=0,\mathrm{x}=0,\mathrm{h}=1$

$$\begin{gathered} \mathrm{F}(0,0)=1 \\ \mathrm{G}(0,0)=1.84147 \end{gathered}$$

$$\begin{gathered} {\mathrm{t}}_{\mathrm{n}+1}=({\mathrm{t}}_{\mathrm{n}}+\mathrm{h}) \\ {\mathrm{x}}_{\mathrm{n}+1}={\mathrm{x}}_{\mathrm{n}}+\frac{\mathrm{h}}{2}(\mathrm{F}+\mathrm{G}) \\ \mathrm{t}=1 \\ \mathrm{x}=1.42074 \end{gathered}$$

Hence $\mathrm{Ï•}\left(1\right)=\mathrm{y}(1,1)=1.42074$

$\mathrm{t}=0,\mathrm{x}=0,\mathrm{h}=0.5$

$$\begin{gathered} \mathrm{F}(0,0)=1 \\ \mathrm{G}(0,0)=1.1237 \end{gathered}$$

$$\begin{gathered} \mathrm{t}=0+0.5 \\ =0.5 \\ \mathrm{x}=0+0.25(1+1.1237) \\ =0530925 \\ \mathrm{Ï•}(0.5)=0.530925 \end{gathered}$$

$\mathrm{t}=0.5,\mathrm{x}=0.530925,\mathrm{h}=0.5$

$$\begin{gathered} \mathrm{F}(0.5,0.530925)=1.13118 \\ \mathrm{G}(0.5,0.530925)=1.88962 \end{gathered}$$

$$\begin{gathered} \mathrm{t}=0.5+0.5 \\ =1 \\ \mathrm{x}=0.530925+0.25(1.13118+1.88962) \\ =1.28613 \\ \mathrm{Ï•}(1)=\mathrm{y}(1,0.5)=1.28613 \end{gathered}$$

Apply the same procedure for all other values and the values are

$$\begin{gathered} \mathrm{Ï•}(1)=\mathrm{y}(1,0.25)=1.26026 \\ \mathrm{Ï•}(1)=\mathrm{y}(1,0.125)=1.25495 \end{gathered}$$

Since,

$$\begin{gathered} \left|\mathrm{y}(1,0.125)-\mathrm{y}(1,0.25)\right|=\left|1.25495-1.26026\right| \\ =0.00531<0.01 \\ \mathrm{Ï•}\left(\mathrm{t}\right)=1.25495 \end{gathered}$$

Hence the solution is role="math" localid="1664306405053" $\mathit{Ï•}\mathbf{\text{(1) = 1.25495}}$